The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm.
Question. The inner diameter of a cylindrical wooden pipe is $24 \mathrm{~cm}$ and its outer diameter is $28 \mathrm{~cm}$. The length of the pipe is $35 \mathrm{~cm}$. Find the mass of the pipe, if $1 \mathrm{~cm}^{3}$ of wood has a mass of $0.6 \mathrm{~g} .\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$


Solution:

Inner radius $\left(r_{1}\right)$ of cylindrical pipe $=\left(\frac{24}{2}\right) \mathrm{cm}=12 \mathrm{~cm}$

Outer radius $\left(r_{2}\right)$ of cylindrical pipe $=\left(\frac{28}{2}\right) \mathrm{cm}=14 \mathrm{~cm}$

Height $(h)$ of pipe $=$ Length of pipe $=35 \mathrm{~cm}$

Volume of pipe $=\pi\left(r_{2}^{2}-r_{1}^{2}\right) h$

$=\left[\frac{22}{7} \times\left(14^{2}-12^{2}\right) \times 35\right] \mathrm{cm}^{3}$

$=110 \times 52 \mathrm{~cm}^{3}$

$=5720 \mathrm{~cm}^{3}$

Mass of $1 \mathrm{~cm}^{3}$ wood $=0.6 \mathrm{~g}$

Mass of $5720 \mathrm{~cm}^{3}$ wood $=(5720 \times 0.6) \mathrm{g}$

$=3432 \mathrm{~g}$

$=3.432 \mathrm{~kg}$
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